Consider a (smooth) bundle E→_B_, and a (smooth) function f: E → R on the total space. Then it makes sense to talk about the derivatives of f along the fibers. Let C be the subspace of E consisting of all points for which all fiber-wise derivatives of f vanish, so that upon intersecting with any fiber C consists of the critical points of the restriction of f to the fiber. If the fiber is n-dimensional, then C is carved out by n equations, and so generically has codimension n in E.

Let's say that c is a point in C so that the second derivative of f in the fiber is nondenegerate (i.e. has non-degenerate Hessian; i.e. f restricts to a Morse function on the fiber through c). Does it follow that the projection C→_B_ is a local diffeomorphism near c?

The answer is yes when everything is finite-dimensional (and I believe the statement is iff). I am interested in the case when B is finite-dimensional but the fibers of E are infinite-dimensional.

Edit: This is a response to Andrew's question below (since answering in the comments proves difficult).

I'm using the word "Morse" loosely, largely because I don't actually know Morse theory. I would suggest that the definition I give is better than what's traditionally used. What I actually mean is this:

Let M be a smooth manifold and f:M→R a smooth map. What type of object is the second derivative f(2)? In general, you should not talk about it by itself, because it does not transform as a tensor, although the pair (f(1),f(2)) is a vector in the 2-jet bundle over M. But if c is a critical point of f, then f(2)(c) is naturally a symmetric bilinear form TcM x TcM → R. Thus it is a map TcM→T*cM. All I ask is that this map have zero kernel.

But if this condition is too weak, whereas a reasonable stronger condition works, I'd love to hear it.

Can you clarify one thing: you are talking about infinite dimensional fibres and also talking about Hessians and Morse functions. I would expect that to mean that you are thinking of the fibres as Hilbertian manifolds, am I correct? Or do you want more generality?
–
Loop SpaceOct 6 '09 at 6:36

Theo also asked this in the 20-questions seminar at Berkeley, so I'm tagging it as such.
–
Scott Morrison♦Oct 6 '09 at 20:52

2 Answers
2

The part I'm still hesitant about is that the manifold you call C is n-dimensional. Codimension arguments in the infinite dimensional setting are always a little sticky. So, I'm just going to treat it as an assumption.

Then the answer to your question is yes, and doesn't depend on what type of infinite dimensional vector spaces you use as your local model. Let's see why.

First you have a space E which is a fiber bundle over B, via the projection p. Then at any point x in E we have the sequence of vector spaces,

TxF = ker dp --> Tx E ----> Tp(x)B

You also have a map f: E --> R, and hence a map df: TxE --> R = Tf(x)R for all points x in E.

C is defined to be the points in E such that the restriction of df to TxF vanishes. By assumption, C is an n-dimensional manifold.

Now let's talk about second derivatives. There are two problems with second derivatives. The first arrises because we are working in the infinite dimensional setting. This means that we might not be able to think of the second derivative as a bilinear form due to possible convergence problems. However, as you pointed out, we can think of it as a map,

TxE --> TxE*

Now depending on the type of infinite dimensional manifold you are considering, this dual space may take on different meanings (banach dual, Frechet dual, etc). You at least get something in the algebraic dual (of possible non-continuous functionals).

The second problem with second derivatives is that they are usually not defined independent of a choice of coordinates. Again, as you pointed out, the pair (df, second der of f) transforms as a section of the second jet bundle, i.e. the second derivative doesn't change like a tensor but in an affine fashion depending on df.

This means that in general there is no intrinsic way to say the second derivative is "non-degenerate" at a random point of E. However, along C, a portion of the second derivative is still well defined, independently of coordinate choices. It is not the whole second derivative, but just the composite,

H: TcE --> TcE* --> TcF*

(Here "H" is for Hessian or some such thing). Under coordinate changes, H transforms like a tensor because, along C, the restriction of df to TcF vanishes.

Here is the key fact: The tangent space of C is the kernel of H. This is an easy calculation, which can be done in local coordinates.

Finally, we are supposed look at the points of C and we are supposed to consider the restriction of H to TcF. We want to consider a point c in C, where the kernel of this restriction is zero.

But this condition is equivalent to saying that the intersection of TcF with the kernel of H is zero, i.e.

(**) At the point c, TcF has zero intersection with TcC.

Now since TcF is the kernel of dp, this means the the restriction of dp to TcC also has zero kernel, and hence for dimension reasons is an isomorphism. Hence at the point c in C, under your assumptions, the map f: C --> B is a local diffeomorphism.

Great, although I had not intended to make any assumptions about dimension on C. In the first picture, I was simply trying to motivate the question without being able to draw a picture. Does the question fail in general?
–
Theo Johnson-FreydOct 12 '09 at 15:53

Ahh. No. If I had thought about this a little longer, I wouldn't have been worried about your assumption. As you can see the exact same argument shows that dp: T<sub>c</sub>C --> T<sub>p(c)</sub>B must be injective. So the only question is if it can have dimension less than B. Again, by looking at H, you can see that the tangent space to C must have dimension at least n. btw: Your question is completely local, so the fact that you have a fiber bundle is a red herring. As long as your notion of "bundle" includes local trivializations, you can reduce to the case that E is a trivial bundle.
–
Chris Schommer-PriesOct 12 '09 at 21:13

Yes, perhaps I should have asked it locally. I actually do everything in trivializations, but I'm trained in physics :P Thanks for the answer!
–
Theo Johnson-FreydOct 14 '09 at 23:52

For completeness, I should post a follow-up to Chris's answer above. Chris's argument shows that the map dp: TcC → Tp(c)B is an injection, where p is (the restriction of) the projection E → B. But the conditions I originally proposed do not force this to be a surjection except in the finite-dimensional case (when a (co)dimension count shows that dim C is at least dim B), as the following example shows.

On the other hand, let's pick a one-form b and a function c on N, and let's suppose that the exterior derivative d_b_ is nondegenerate, so that d_b_ is a symplectic structure on N. Let L(v,q) = _b_v + c. Then one can check (it's straightforward) that the Euler-Lagrange equations are a non-degenerate first order differential equation on N, equivalent to the Hamilton equation for the sympectic manifold (N,d_b_) with Hamiltonian c (or perhaps -c depending on your conventions). Thus a solution is determined by its initial location, and so dim C = dim N.

On the other hand, the Hessian H defined in Chris's answer is now a nondegenerate first-order linear differential equation, and the condition says that the only solution φ to this equation with φ(0) = 0 = φ(1) is the trivial solution φ = 0. For a general Lagrangian, the Hessian is a second-order operator, and this condition is nontrivial, but when the Lagrangian is first-order, the Hessian necessarily has no kernel — a solution is determined by a single value.

Thus the condition that Chris thought I was imposing — that C be a manifold with dimension the same as B — cannot be dropped.

A final remark is that in finite dimensions, C is cut out by dim F = dim E - dim B equations, and so dim C is at least dim B. The point is that this dimension count fails when dim F = ∞.